Forecasting project (Meandering pattern)

PA 453 - Forecasting Project

(CASE)

You are responsible for investing your agency's liquid cash. You are considering investing in a particular security portfolio that has yielded the following monthly returns. How would you predict its behavior for the next three months? Justify your choice of method. (Reference: PA453S01Assignments.doc, Lilliard Richardson, Graduate School of Public Affairs, University of Missouri Columbia, MO 65211-2015)

(Figure 1) drawing the graph at the beginning

Pattern of the data

Meandering pattern, which does not contain any systematic or repeating pattern

(APPROACH)

Two ways in forecasting analysis on the meandering pattern

1. Trend analysis

2. Autoregressive modeling

Which forecasting ways I choose will fundamentally depend on policy goal/objectives and environment (such as time, resources and purpose). Herein I will try to carry out both ways.

Trend analysis

I will carry out trend analysis to detect the underlying pattern in the data.

(Procedure)

1. Creating regression model

2. Check the model’s figure and R² in each regression model

3. Residual analysis for determining residual values

4. Choosing best model and put independent variables for forecasting

Creating regression models

	Regression model	Standard error of the estimate
Linear	Y=0.2529x+2.6705	R²=0.1251
Logarithmic	Y=0.767LN(x)+3.2668	R²=0.0352
Polynomial	Y=0.0685x²-0.8427x+5.7745	R²=0.2604
Power curve	Y=1.973x^0.2224	R²=0.0185
Exponential curve	Y=1.4625e^0.0891x	R²=0.0971

Selection best curve for forecasting yield of the next three months

Selection criteria 1: we choose the curve with the highest R²

Under the criteria 1, I can choose polynomial curve (R²=0.2604).

Selection criteria 2: minimum MAPE (we select the curve that has smallest error over the most recent several months)

The result from computing the Mean Absolute Percentage Error (MAPE) is as following.

(Figure2) Calculation of MAPE

	Linear	Logarithmic	Polynomial	Power	Exponential
MAPE	13.60%	26.50%	15.66%	50.70%	28.87%

Rationale of choosing liner regression model

For example, over the most recent three months, on average, the predicted liquid cash based on the liner model differed from the actual liquid cash by 13.60%. In conclusion, based on the second criteria, liner regression model is best among the five models.

I therefore choose the liner model for two reasons. First, its MAPE was lowest value. In addition, I reject the polynomial model because it indicated that liquid cash had peaked and was going down. This situation is not consistent.

(Figure 3) result of forecasting on the basis of polynomial model

Residual analysis

Before I use the liner model for forecasting, I should conduct a residual analysis to determine if the model satisfies residual value. The residual plot for a model that satisfies the statistical independence assumption should be horizontal band of residual centered at zero that contain no pattern. Figure 5 suggests there is no pattern to the residuals over time.

(Figure 4) residual analysis

(Figure 5) drawing residual graph

I can use liner model for short-term forecasting. The forecasting result is as following.

Autoregressive modeling

On the other, I will check the result of autoregressive model approach by using Excel data-analysis regression.

(The Regression Equation)

“The regression equation is the equation for the straight line that best describes a set of data point. The equation is asymmetric; therefore, the independent and dependent variables must be designated before calculating the equation” (Sullivan & Rassel, 429).

Y = a + b*x

Y= dependent variable (in this case, yield (t-1))

a= intercept or constant

b= regression coefficient for X associated with Y

x= independent variable (in this case, yield (t))

The regression equation gives the independent effect of x (yield t-1) over increase (or decrease) of yield t in this case. (Articulating relationships between yield t and yield t-1 during study period)

(Procedure of forecasting in the autoregressive model)

1. Getting data on two interval variables

2. Plotting the data on a scatter graph (the dependent variable Y is graphed on the vertical axis while the independent variable x is graphed on the horizontal axis.)

3. Deciding whether the plotted data can be appropriately accommodate in a straight line

4. Calculating the regression equation

5. Doing Analysis of Variance (ANOVA) to make the correlation clear

6. Putting data into the regression equation for forecasting implementation

(Figure 6) assignment of two variables (yield t and yield t-1)

(Figure 7) drawing a scatter graph (It looks like there will be no statistically clear relationship between two variables)

(Figure 8) ANOVA

Figure 7 is the scatter plot of yield t for a period versus yield t-1 for the previous period for month 2-15. The swarm of the 15 data points is neither upward nor downward. That is the figure is scattered. It suggest that yield t and yield t-1 are not positively and linearly related. In addition, From the ANOVA table (figure 8), the p-value for the model is 0.8804. In terms of hypothesis testing (null= there is no relationship between yield t and yield t-1), 0.8804 is bigger than 0.05 % significant level. So, I cannot reject the null. Consequently, there is no possibility to use data-analysis regression tool to fit a liner regression model to this case.

In conclusion, I will recommend the forecasting data based on trend analysis, but not autoregressive analysis.